# Moment of Inertia of Cavendish pendulum

1. Aug 19, 2007

### Pioneer1

Hi,

I am trying to compute the moment of inertia of the Cavendish pendulum. I used

I = 2(m r^2)

r = gyration arm
m = the weight attached to the pendulum

But this formula is for a dumbell type of torsion pendulum where the weights are attached to the bar.

Does anyone know the formula for a pendulum where the weights are suspended as in the case of the Cavendish pendulum? I couldn't find it online.

With this formula I got

I = 2 ( 729.8 * 93.1^2) = 12,651,243.56 g cm^2

Does this sound right for a pendulum of this dimensions?

2. Aug 19, 2007

### Staff: Mentor

For the benefit of those of us who haven't been following your previous discussions closely, can you provide a diagram of the specific geometry that you have in mind, so we can see how it differs from Example 1 in the link you provided (which is the case where $I = 2mr^2$)?

3. Aug 19, 2007

### Pioneer1

http://www.alphysics.com/cavendishexperiment/CavendishSchematic11.jpg

And here's a detail

http://www.alphysics.com/cavendishexperiment/CavendishSchematic111.jpg

Cavendish also gave the weight of the bar and the silver wire he used to strengthen it. I assume those will have to be taken into consideration as well.

4. Aug 19, 2007

### Staff: Mentor

You've still got two masses revolving around a common central point, so their moment of inertia I is the same as when they're connected by a rod of negligible mass in a dumbbell configuration as in the page you first referenced. So you're OK so far.

What you need now is I for the rotating part of the suspension mechanism. You have the mass of the transverse rod, and I for a rod rotating transversely around its midpoint is a standard case. For some reason it's not on the page you first referenced, but you can find it on Hyperphysics:

http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html#cmi

Just add the two I's together. In principle, you should also consider I for the rest of the rotating part of the apparatus, but the supporting wires are probably pretty light so you can safely neglect them, at least as a start in your analysis.

5. Aug 19, 2007

### Pioneer1

Ok. Thanks. From hyperphysics I got 1/12 M L^2 and added it to the previous result:

I = 2 m r^2 + 1/12 M L^2

L = length of the rod
r = half length of the rod

So,

Total moment of inertia for the Cavendish experiment = r^2 (2m + 1/3M) = 13,137,851.84 cm2 g

m = weight of each ball = 729.8 g

M = Rod weight = 155.5 + 11 + 2.9 = 169.40 g (including deal rod, silver supporting wire, vernier attached to the arm)

L = Rod length = 73.3 inches = 186.18 cm

r = Half rod length = 93.09 cm

I also added another picture with a little more detail about the arm. (Figure 3 here.)

Does this look good?